Hmwk: pg 204 #1-2(ace), 5, 6,7ace,9ac, 13, 15, (16), (18)
Solving Polynomial Equations Algebraically
Recall:
Prior knowledge suggests that a polynomial equation of degree two can be
solved using the quadratic formula or a factoring method.
Example 1
Solve the following equation using two methods.
6x2 +15x = 9
Method 1 (Quadratic Formula)
Method 2 (Factoring)
Notice that both methods require that a ______ be put on one side of the
equation to solve it.
The second method (factoring) makes use of the Zero Principle which states
that if any factor in the factored expression can be made to be zero by
assigning some value to x then the entire factored expression becomes zero
and this value of x is a solution.
The Zero Principle
If a polynomial equation can be written as a product as follows:
k(x - a1)(x - a2)(x - a3)....(x - an) = 0 where k, a1, a2, a3,..., an are all constants,
then the solutions (or roots) to the equation are a1, a2, a3,..., an.
The Zero Principle can also be used to solve one variable polynomial
equations of degree larger than 2.
Example 2
Solve each equation using factoring techniques.
a) x3 - 4x2 - 5x = 0
b) 4 x3  12 x 2  x  3  0
c) x3 – 5x = 2x2 – 6
d) x3 – 4x2 + 8x – 8 = 0
e) x4 + 4 = 5x2
f) x4 + 2x3 + 14x2 + 13x + 36 = 0